# Mass to fuel ratio of a hypothetical spaceship propelled using a helium-3 reactor

Discussion in 'Physics & Math' started by s0meguy, Jul 19, 2013.

1. ### s0meguyWorship me or suffer eternallyValued Senior Member

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Hey there. I'm trying to get a good idea of said mass to fuel ratio. Is the energy that can be extracted from Helium-3 in a nuclear reactor much higher than that of the oil-based fuels with the same mass that we use on Earth? Assuming that the ship has to be able to accelerate at 1.5g for 1 year and ignoring all other energy needs of the crew. Since we're talking about a ratio here I guess we don't need the ship's mass but if you do it is 10,000,000 kg. I am not saying that it is constantly accelerating, it needs to have enough fuel to be able to do so, though.

How much mass in helium-3 is needed to do that assuming 100% conversion of energy into propulsion?

With my limited understanding I have made this calculation in the case of hydrogen atom antimatter annihilation, please tell me if I am correct and if not please say what I did wrong or provide the correct calculation.

Thanks a lot for helping.

3. ### Fednis48Registered Senior Member

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725
Looks right, at least at first glance. If you want to simplify the calculation, you don't need to find the energy of each helium atom and then multiply by the number of atoms to get total energy; you can just plug the total mass of helium in as $m$ in $E=mc^2$. If you also just want the ratio of masses, you can make it even simpler:

$W=fs=m_{fuel}c^2$

$m_{ship}as=m_{fuel}c^2$

$\frac{m_{fuel}}{m_{ship}}=\frac{as}{c^2}$

As a rule of thumb, you'll get cleaner results if you do everything symbolically as far as you can, then plug in numbers at the very end.

5. ### Q-reeusBannedValued Senior Member

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4,695
You really can't ignore the contribution of a continually changing fuel mass. May I suggest a read here:
http://www.relativitycalculator.com/relativistic_photon_rocket.shtml

7. ### Janus58Valued Senior Member

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2,301
1 year ship time or Earth time?.

If ship time, you can use

$v= c \tanh \left ( frac{aT}{c} \right )$

to find your final velocity. It works out to 0.918c

The KE can be found by

$KE= mc^2 \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right )$

With m the mass of the ship. We are looking for M the mass needed to generate that much energy via total conversion,so

$M= m \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right )$

which gives 15.2 million kg. ( upon checking your figures, you shifted a decimal point in your last calculation, you should have gotten 12.4 million kg, not 124 million.)

Now, if we assume a photonic rocket (one which uses photons as its exhaust.), we can account for the mass of our fuel and use

$\frac{M}{m}= e^{\frac{aT}{c}}-1$

which works out to 3.84, which would give a M of 38.4 million kg.

We can also use this last formula to answer the original question of using Helium-3 in a reactor.

A He3-He3 reaction liberates 12.86 Mev. Now this is not total conversion, so we are left with He4 + two protons, which we use as our reaction mass. This give an exhaust velocity of the order of 0.016c

Plugging this into our formula gives:

$\frac{M}{m}= e^{\frac{aT}{0.016c}}-1$

M/m = 6.3e42

In other words, you would need 2.5 times the mass of the galaxy in He3 to accelerate a payload 1kg for 1 year at 1.5g using a He3-He3 fusion rocket.

If you are considering 1 year Earth time, then use 0.784 for T.

This would have the effect of dropping the final answer to only needing 1794 solar masses of fuel and the final velocity to 0.844c

8. ### s0meguyWorship me or suffer eternallyValued Senior Member

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Thanks a lot guys. I thought that Helium-3 was more viable as a long distance propulsion fuel but with those numbers it doesn't sound like it. The fuel to mass ratio is just so huge. Sounds like our only real fuel option is antimatter.